Tct =给定类中术语的出现次数
Tct’是给定类中的#术语
B’=所有文件中的#唯一术语
classes =垃圾邮件或火腿
spam = spam,ham = not spam
问题是,当我在我的代码中使用这个公式时,它给了我非常小的条件概率分数,如2.461114392596968e-05.我很确定这是因为Tct的值非常小(如5或8),而Tct’的分母值(火腿为64878,垃圾邮件为308930)和B'(为16386).我无法弄清楚如何将condprob分数降低到像.00034155这样的东西,因为我只能假设我的condprob分数不应该像它们那样指数级小.我的计算错了吗？这些值实际上应该是这么小吗？
如果它有帮助,我的目标是为一组测试文档打分并获得如327.82,758.80或138.66的结果
使用这个公式
但是,使用我的小condprob值我只得到负数.
码
– 创建文档
class document(object):"""The instance variables are:filename....The path of the file for this document.label.......The true class label ('spam' or 'ham'),determined by whether the filename contains the string 'spmsg'tokens......A List of token strings."""def __init__(self,filename=None,label=None,tokens=None): """ Initialize a document either from a file,in which case the label comes from the file name,or from specifIEd label and tokens,but not both. """ if label: # specify from label/tokens,for testing. self.label = label self.tokens = tokens else: # specify from file. self.filename = filename self.label = 'spam' if 'spmsg' in filename else 'ham' self.tokenize()def tokenize(self): self.tokens = ' '.join(open(self.filename).readlines()).split() -NaiveBayes
class NaiveBayes(object):def train(self,documents): """ Given a List of labeled document objects,compute the class priors and word conditional probabilitIEs,following figure 13.2 of your book. Store these as instance variables,to be used by the classify method subsequently. Params: documents...A List of training documents. Returns: nothing. """ ###Todo unique = [] proxy = [] proxy2 = [] proxy3 = [] condprob = [{},{}] Tct = defaultdict() Tc_t = defaultdict() prior = {} count = 0 oldterms = [] old_terms = [] for a in range(len(documents)): done = False for item in documents[a].tokens: if item not in unique: unique.append(item) if documents[a].label == "ham": proxy2.append(item) if done == False: count += 1 elif documents[a].label == "spam": proxy3.append(item) done = True V = unique N = len(documents) print("N:",N) LB = len(unique) print("THIS IS LB:",LB) self.V = V print("THIS IS COUNT/NC",count) Nc = count prior["ham"] = Nc / N self.prior = prior Nc = len(documents) - count print("THIS IS SPAM COUNT/NC",Nc) prior["spam"] = Nc / N self.prior = prior text2 = proxy2 text3 = proxy3 TctTotal = len(text2) Tc_tTotal = len(text3) print("THIS IS TCTOTAL",TctTotal) print("THIS IS TC_TTOTAL",Tc_tTotal) for term in text2: if term not in oldterms: Tct[term] = text2.count(term) oldterms.append(term) for term in text3: if term not in old_terms: Tc_t[term] = text3.count(term) old_terms.append(term) for term in V: if term in text2: condprob[0].update({term: (Tct[term] + 1) / (TctTotal + LB)}) if term in text3: condprob[1].update({term: (Tc_t[term] + 1) / (Tc_tTotal + LB)}) print("This is condprob",condprob) self.condprob = condprobdef classify(self,documents): """ Return a List of strings,either 'spam' or 'ham',for each document. Params: documents....A List of document objects to be classifIEd. Returns: A List of label strings corresponding to the predictions for each document. """ ###Todo #return List["string1","string2","stringn"] # docs2 = ham,condprob[0] is ham # docs3 = spam,condprob[1] is spam unique = [] ans = [] hscore = 0 sscore = 0 for a in range(len(documents)): for item in documents[a].tokens: if item not in unique: unique.append(item) W = unique hscore = math.log(float(self.prior['ham'])) sscore = math.log(float(self.prior['spam'])) for t in W: try: hscore += math.log(self.condprob[0][t]) except KeyError: continue try: sscore += math.log(self.condprob[1][t]) except KeyError: continue print("THIS IS Sscore",sscore) print("THIS IS Hscore",hscore) unique = [] if hscore > sscore: str = "Spam" elif sscore > hscore: str = "Ham" ans.append(str) return ans -测试
if not os.path.exists('train'): # download datafrom urllib.request import urlretrIEveimport tarfileurlretrIEve('http://cs.iit.edu/~culotta/cs429/lingspam.tgz','lingspam.tgz')tar = tarfile.open('lingspam.tgz')tar.extractall()tar.close()train_docs = [document(filename=f) for f in glob.glob("train/*.txt")]test_docs = [document(filename=f) for f in glob.glob("test/*.txt")]test = train_docsnb = NaiveBayes()nb.train(train_docs[1500:])#uncomment when testing classify()#predictions = nb.classify(test_docs[:200])#print("PREDICTIONS",predictions) 最终目标是能够将文档分类为垃圾邮件或火腿,但我想首先处理条件概率问题.
问题
条件概率值应该是这么小吗？如果是这样,为什么我通过分类获得奇怪的分数？如果没有,我如何修复我的代码以给我正确的condprob值？
值
我得到的当前condprob值是这样的：
‘Tradition’：2.461114392596968e-05,’fillmore’：2.461114392596968e-05,’796’：2.461114392596968e-05,’zann’：2.461114392596968e-05
condprob是一个包含两个词典的列表,第一个是火腿,第二个是垃圾邮件.每个字典都将一个术语映射到它的条件概率.我希望有“正常”的小值,例如.00031235而不是3.1235e-05.
这样做的原因是,当我通过classify方法运行condprob值和一些测试文档时,得到的分数就像
这是Hscore -2634.5292392650663,这是Sscore -1707.983339196181
什么时候应该看起来像
这是Hscore 327.82,这是Sscore 758.80
运行时间
~1分30秒
从我的代码中可以看出,你先从先验概率p(Ham)和p(Spam)开始,然后使用从先前数据估计的概率来计算p(Ham)* p(观测数据| Ham)和p(垃圾邮件) )* p(观察数据|垃圾邮件).
贝叶斯定理重新排列p(Obs | Spam)= p(Obs& Spam)/ p(Spam)= p(Obs)p(Spam | Obs)/ p(Spam)给你P(Spam | Obs)= p(垃圾邮件)p(Obs | Spam)/ p(Obs)你似乎计算了p(垃圾邮件)p(Obs | Spam)= p(Obs& Spam)但没有除以p(Obs).由于Ham和Spam只有两种可能性,最简单的做法可能是注意p(Obs)= p(Obs& Spam)p(Obs& Ham),所以只需将两个计算值中的每一个分开通过它们的总和,基本上缩放值,使它们确实总和为1.0.
如果从日志概率lA和lB开始,这种缩放比较棘手.为了缩放这些,我首先将它们通过粗略值作为对数来缩放它们,所以做一个减法
lA = lA – max(lA,lB)
lB = lB – max(lA,lB)
现在至少两者中较大者不会溢出.较小的仍然可能,但我宁愿处理下溢而不是溢出.现在将它们变成不完全缩放的概率：
pA = exp(lA)
pB = exp(lB)
并适当缩放,使它们增加到零
truePA = pA /(pA pB)
truePB = pB /(pA pB)
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